Suppose
100,000 newborn mice are produced by a population in
which the frequency of allele A = 0.3. The newborns
are distributed according to Hardy-Weinberg expectations,
such that there are 9,000 AA, 42,000 AB, and
49,000 BB mice. Suppose the fitness of the three
types, measured as viability (the
probability of survival to reproductive age), of each
of the three phenotypes corresponding to genotypes AA,
AB, and BB is 0.5, 0.4, and 0.3,
respectively. This is an additive model,
in which each B allele decreases viability by 0.1.
Therefore, the expected numbers of survivors to reproductive
age are:
AA: 0.5 x 9,000
newborns = 4,500 AA adults
AB: 0.4 x 42,000 newborns = 16,800
AB adults
BB: 0.3 x 49,000 newborns = 14,700
BB adults
The total number of
surviving adults is (4,500 + 16,800 + 14700) = 36,000.
Following established principles, f(A) = [(2)(4,500) +
16,800] / (2)(36,000) = 0.3583, and f(B) =
[(2)(14,700) + 16,800] / (2)(36,000) = 0.6417 = 1 - f(A).
Viability selection has therefore decreased the relative
frequency f(B)' by 0.7000 - 0.6417 = 0.0583,
about 6%.
In this case, evolution defined as changes
in allele frequencies, occurs in a population that declines
in numbers between the parents and offspring. Allele
frequency change is determined from relative rather
than absolute viability. With more intense selection,
if relative viabilities were 0.05, 0.04, and 0.03,
the total number of mice surviving to adulthood would be only
3,600, but the calculated allele frequency change would be the
same as before [HOMEWORK:
Prove this].
Alternatively, viability selection
may occur while the population size remains constant,
if the expected number of newborn mice of each
genotype surviving to adulthood is proportional to the
carrying capacity of the environment, in this case 100,000.
Expected numbers can be predicted by weighting by
the ratio of newborns to that of the adults, in this case
100,000/36,000 = 2.778. Then
AA: 0.5 x 9,000
newborns x 2.778 = 12,500 AA adults
AB: 0.4 x 42,000 newborns x
2.778 = 46,670 AB adults
BB: 0.3 x 49,000 newborns x 2.778
= 40,830 BB adults
The expected total
number of newborns that survive to adulthood is now 100,000.
Calculated allele frequency change would be the same as in
the original model [HOMEWORK:
Prove this]. This is essentially K-selection
that acts so as to maintain N near a constant
carrying capacity K. Computer models (such as NatSel)
simulate this by generating genotypes at random, applying
the viability of each type, and continuing to add to the
population in this way, until the total number of
individuals again reaches the carrying capacity of
100,000.
Finally,
note that weighting by a larger factor of (say
3.0) would result in an increase in the
population size, but once again calculated allele
frequency changes would be the same. Although population
size increases, this would not be due to an
increase in the frequency of the allele that offers the
greater fitness advantage. This again shows that relative Darwinian fitness
and absolute survival
are unrelated.